How do you evaluate the function with the given values of $x: color(white)("d")f(x)=(1/2)[x+1] ;color(white)("ddd") x=0; x=-1$ ?

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Answer 1 · 2018-06-22T15:13:13

$f(0)->y=1/2$
$f(-1)->y=0$

Suppose $x=0$ then wherever there is an $x$ write 0

$f(x)=1/2(x+1) color(white)("ddd")-> color(white)("ddd")f(0)=1/2(0+1)=1/2$

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Suppose $x=-1$ then wherever there is an $x$ write -1

$f(x)=1/2(x+1) color(white)("ddd")-> color(white)("ddd")f(0)=1/2[color(white)(2/2)(-1)+1color(white)("d")]=0$

How do you evaluate the function with the given values of x: color(white)("d")f(x)=(1/2)[x+1] ;color(white)("ddd") x=0; x=-1x: color(white)("d")f(x)=(1/2)[x+1] ;color(white)("ddd") x=0; x=-1?